# Motion of a charged particle in a constant and uniform magnetic field

Assuming the following relationship has been demonstrated

$$r=\frac{m u_{0}\sin \theta_0}{qB\sqrt{1-\left(\dfrac{u_{0}^2}{c^2}\right)}}=\frac{p_0\sin \theta_0}{qB}$$

where $$p_0=mu_0/\sqrt{1-\beta^2}$$ represents the relativistic momentum of the particle. In the relativistic case, therefore, $$\omega$$ (relativistic angular velocity $$\omega=\frac{qB}{m} \sqrt{1-\left(\frac{u_{0}^2}{c^2}\right)^2}$$) is no longer constant but depends on the speed $$\bar{u}_{0}$$ of the charged particle $$q$$; in fact the factor $$\gamma$$ is present.

The step $$p$$ (different of the momentum $$p_0$$) is obtained from the product of the parallel component of the initial particle speed $$u_{0\parallel}$$ and the period $$T$$:

$$p=(u_{0})_zT=u_{0\parallel}T$$

Making some considerations about the angle $$\theta_0$$, if $$\theta_0=0$$ why I obtain a straight line? Don't you have $$r=0$$?

• Why do some users want to close my question? Is there a reason? – Sebastiano Jul 12 '19 at 22:21
• It isn’t clear what $r$ is the radius of, what $\theta_0$ is the angle between, what $\omega$ is supposed to be, and how $\bar{u}$, $u_0$, and $u$ differ. – G. Smith Jul 12 '19 at 22:27
• Also, in relativity, $u_0$ and $p_0$ usually mean the covariant time component of the four-momentum, but I don’t think you’re using them that way. – G. Smith Jul 12 '19 at 22:30
• Finally, I don’t think you mean “$\omega$ is no longer constant”. You seem to mean “$\omega$ is no longer speed-independent”. – G. Smith Jul 12 '19 at 22:32
• So, overall, the question is quite unclear. – G. Smith Jul 12 '19 at 22:35

When $$\theta_0$$ is zero, the trajectory is no longer a spiral around the $$z$$-axis; it is a straight line along the $$z$$-axis, for which $$r=0$$ in cylindrical coordinates.